tag:blogger.com,1999:blog-6933544261975483399.post4157505329491254372..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1186: Right Triangle, Square, Inscribed Circle, Tangent, Quadrilateral, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-43780434581394778442016-02-09T02:52:36.771-08:002016-02-09T02:52:36.771-08:00MANOLOUDIS APOSTOLIS
4 HIGH SCHOOL OF KORYDALLOS -...MANOLOUDIS APOSTOLIS<br />4 HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE<br />PROBLEM 1186<br /> <br />Suffice 2.(BHG)=(ABC) or 2.(BG.BH)/2=(AB.BC)/2 or 2.BG.BH=AB.BC.Formed square BCFS end externally tangent circle with center K eat in accordance with the problem 1184 PT <br />Tangent to the circle with center K. Fetch and second tangent from P intersect the AS to M and DB to L.Is BG=BM(symmetrical with respect to PO).Fetch OQ perpendicular to AB. Is triangles OTH=OHQ Therefore <TOH= <QOH=x. But <BHG=<QOT=2x,<HGB=90-2x.Is <KGP= ½(90+2x)=45=x=<HOK.Therefore OHGK is concycle. Simple OLMK is concycle.But PG.PH=PK.PO=PM.PL therefore O,L,M,K,G,H is concycle.<br />2.BG.BH=2.BM.BH=2.BK.BO=2.(BC.√(2&2))/2.(AB.√2)/2=AB.BC<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28081675755568924372016-02-08T14:36:22.210-08:002016-02-08T14:36:22.210-08:00http://s17.postimg.org/m0fx94qbz/pro_1186.png
Dra...http://s17.postimg.org/m0fx94qbz/pro_1186.png<br /><br />Draw circle I similar to problem 1184<br />Observe that TP tangent to circle O and I<br />Let 2m and 2n are diameters of circles O and I<br />Let u= BH and v= BG, a= HT=HM and b= GN=GR<br /> Applying Menelaus theorem in tri. ABC and secant HGP<br />(BH/HA).(PA/PC).(GC/GB)=1….. (1)<br />Replace PA/PC=AE/BC= m/n in (1) and simplify it we get<br />u.v/(v-u)= 2.m.n/(m-n) ……. (2) <br />Since TP , BC and BA are ext. tangent to circles O and I so we have<br />Tangent from G to circle O GT=GQ => GH+a= b+NQ….. (3)<br />Tangent from H to circle I GT=GQ => GH+b= a+MS….. (4)<br />Subtract (3) to (4) and note that NQ=MS we get a= b. <br /> So HM=HT=GN or v-u= m-n<br /><br />Replace it in (2) we will have u.v=2.m.n => area of triangle HBG= ½ area of triangle ABC<br />So line HG bisect triangle ABC into 2 congruent areas<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com